Question

我有一个用户表和一个消息表。

在消息表（名为messaggi）中，有msg_to和msg_from，他们的消息（msg_text）和msg_date（这是一个日期和时间）。

我试图获得一个列表：

来自utenti表
对于邮件为SENT或RECEIVED的每个用户，最后应显示
隐藏非自我人的信息（在这种情况下为id_utente 1 = self）。

这是我提出的，但我继续获取所有消息或双重用户，等等......

SELECT CONCAT(LEFT(u.fname, 1), LEFT(u.lname, 1)) AS iniziali,
       u.email,
       u.color,
       CONCAT(u.fname, " ", u.lname) AS full_name,
       MAX(m.msg_date) AS msg_date,
       m.msg_text
FROM utenti u
     INNER JOIN messaggi m ON m.msg_to = u.id_utente
WHERE m.msg_to = 1
GROUP BY m.msg_to,
         m.msg_from
UNION
SELECT CONCAT(LEFT(u2.fname, 1), LEFT(u2.lname, 1)) AS iniziali,
       u2.email,
       u2.color,
       CONCAT(u2.fname, " ", u2.lname) AS full_name,
       "",
       ""
FROM utenti u2
WHERE u2.id_utente NOT IN
(
    SELECT id_utente
    FROM utenti u
         INNER JOIN messaggi m ON m.msg_to = u.id_utente
    WHERE m.msg_to = 1
          AND u.id_utente = 1
    GROUP BY m.msg_to,
             m.msg_from
);

这是一个小提琴：

http://sqlfiddle.com/#!9/106319/1

我想要的输出应该是：

| iniziali |       email |  color |    full_name |            msg_date |       msg_text |
|----------|-------------|--------|--------------|---------------------|----------------|
|       BV |  456@me.com | (null) |   Bill Villa | 2018-04-20 12:29:20 | Msg 2 (1 to 2) |
|       MG |  789@me.com | (null) |    Max Gazze | 2018-04-09 14:59:39 | Msg 1 (3 to 1) |
|       JB |  101@me.com | (null) |    Jack Blue |                     |                |

在这种情况下，我是id＃1我只看到从ME和TO ME发送的消息，并且对于每个用户，我只看到最后一个接收或发送的消息。像WhatsApp，Facebook Messanger，Telegram等......你会看到每个人发送/接收的联系人和最后一条消息。

不显示与其他用户（即用户2到用户3）之间的消息。

正如你所看到的，我没有在用户列表中看到自己（id＃1），而对于Jack Blue我只看到他的名字e没有消息，因为没有消息从用户1和4发送过。所以我最后得到一个用户列表，对于每个用户，我都会看到最近的消息（已发送或已接收），但在没有消息的情况下，我只看到空闲msg_text和msg_date

Answer 1

好的，所以我认为我可能有一个解决方案。我用你的小提琴测试它似乎工作正常。

通过组合MySQL GROUP BY行为（仅返回组中的第一个）和ORDER BY，我们可以实现所需的结果，尽管最终结果按名称排序。（编辑：只需将ORDER BY msg_date desc添加到SQL查询的末尾，然后按日期排序）

编辑：我已将查询更改为id_utente上的分组，并按msg_date排序。它适用于小提琴，它正在运行MySQL version 5.6

SELECT * FROM ((SELECT CONCAT(LEFT(u.fname , 1), LEFT(u.lname , 1)) as iniziali, 
        u.id_utente,
        u.email, 
        u.color, 
        CONCAT(u.fname, " ", u.lname) as full_name, 
        m.msg_date,
        m.msg_text 
  FROM utenti u
  LEFT JOIN messaggi m ON m.msg_to = u.id_utente
  WHERE (m.msg_from = 1) OR m.msg_date is NULL)
  UNION
  (SELECT CONCAT(LEFT(u.fname , 1), LEFT(u.lname , 1)) as iniziali, 
        u.id_utente,
        u.email, 
        u.color, 
        CONCAT(u.fname, " ", u.lname) as full_name, 
        m.msg_date,
        m.msg_text 
  FROM utenti u
  LEFT JOIN messaggi m ON m.msg_from = u.id_utente
  WHERE (m.msg_to = 1) OR m.msg_date is NULL)
  ORDER BY id_utente, msg_date desc) st1
  WHERE id_utente != 1
  GROUP BY id_utente ORDER BY msg_date desc

在你的小提琴中，这将返回

iniziali | id_utente |  email      | color  | full_name  | msg_date             | msg_text
---------+-----------+-------------+--------+------------+----------------------+-------------------
BV       |         2 | 456@me.com  | (null) | Bill Villa | 2018-04-20T12:29:20Z | Msg 2 (1 to 2)
MG       |         3 | 789@me.comm | (null) | Max Gazze  | 2018-04-09T14:59:39Z | Msg 1 (3 to 1)
JB       |         4 | 101@me.comm | (null) | Jack Blue  | (null)               | (null)

Answer 2

这是我对此的尝试，我使用了一个联合，第一部分是针对那些来自/到id 1的消息的用户，而第二部分是针对那些没有消息的用户。

SELECT CONCAT(LEFT(u.fname, 1), LEFT(u.lname, 1)) AS iniziali,
   u.email,
   u.color,
   CONCAT(u.fname, " ", u.lname) AS full_name,
   u.id_utente,
   m.msg_date,
   m.msg_text
FROM utenti u, messaggi m
WHERE (m.msg_to = u.id_utente OR m.msg_from = u.id_utente)
AND (m.msg_to = 1 and m.msg_from != 1 OR m.msg_to != 1 and m.msg_from = 1)
AND m.msg_date = (SELECT MAX(m2.msg_date) FROM messaggi m2 WHERE (m2.msg_to = u.id_utente AND m2.msg_from = 1) OR (m2.msg_from = u.id_utente AND m2.msg_to = 1))
UNION ALL
SELECT CONCAT(LEFT(u.fname, 1), LEFT(u.lname, 1)) AS iniziali,
   u.email,
   u.color,
   CONCAT(u.fname, " ", u.lname) AS full_name,
   u.id_utente,
   '',
   ''
FROM utenti u
WHERE NOT EXISTS (SELECT * FROM messaggi m WHERE (m.msg_to = u.id_utente AND m.msg_from = 1) OR (m.msg_from = u.id_utente AND m.msg_to = 1))
AND u.id_utente != 1
ORDER BY msg_date DESC

Answer 3

尽管你已经接受了答案，但我已经做出了努力，我也想发布这篇文章。此查询也将根据您提供的示例数据起作用。我使用了一个变量来测试不同的用户。

set @uid = 1;
select u1.id_utente as id,
    CONCAT(LEFT(u1.fname, 1), LEFT(u1.lname, 1)) AS iniziali,
    u1.email as email,
    u1.color as color,
    concat(u1.fname, ' ', u1.lname) as full_name, 
    m.msg_date as msg_date,
    m.msg_text as msg_text
from utenti u1
join utenti u2 
    on u1.id_utente != u2.id_utente and u2.id_utente = @uid
left join messaggi m
    on m.msg_to = u1.id_utente and m.msg_from = u2.id_utente or
       m.msg_to = u2.id_utente and m.msg_from = u1.id_utente
where m.msg_date = (select max(msg_date) 
                    from messaggi m2
                    where m2.msg_to = u1.id_utente and m2.msg_from = u2.id_utente or
                    m2.msg_to = u2.id_utente and m2.msg_from = u1.id_utente) or
      m.id_msg is null
group by u1.id_utente
order by msg_date desc

@uid = 1，输出

id  iniziali    email       color   full_name       msg_date                msg_text
2   BV          456@me.com  (null)  Bill Villa      2018-04-20T12:29:20Z    Msg 2 (1 to 2)
3   MG          789@me.com  (null)  Max Gazze       2018-04-09T14:59:39Z    Msg 1 (3 to 1)
4   JB          101@me.com  (null)  Jack Blue       (null)                  (null)

@uid = 2，输出

id  iniziali    email       color   full_name       msg_date                msg_text
1   JL          123@me.com  (null)  Joe Lombardi    2018-04-20T12:29:20Z    Msg 2 (1 to 2)
3   MG          789@me.com  (null)  Max Gazze       2018-04-09T15:03:44Z    Msg 1 (3 to 2)
4   JB          101@me.com  (null)  Jack Blue       (null)                  (null)

@uid = 3，输出

id  iniziali    email       color   full_name       msg_date                msg_text
2   BV          456@me.com  (null)  Bill Villa      2018-04-09T15:03:44Z    Msg 1 (3 to 2)
1   JL          123@me.com  (null)  Joe Lombardi    2018-04-09T14:59:39Z    Msg 1 (3 to 1)
4   JB          101@me.com  (null)  Jack Blue       (null)                  (null)

@uid = 4，输出

id  iniziali    email       color   full_name       msg_date                msg_text
1   JL          123@me.com  (null)  Joe Lombardi    (null)                  (null)
2   BV          456@me.com  (null)  Bill Villa      (null)                  (null)
3   MG          789@me.com  (null)  Max Gazze       (null)                  (null)

用户的MSYQL SELECT和群组对话并显示上一条消息

3 个答案: